Prove that, if S is open, each of its points is a point of accumulation of S.
How can I prove this??

January 20th 2011, 05:54 PM

Drexel28

Quote:

Originally Posted by mathsohard

Prove that, if S is open, each of its points is a point of accumulation of S.
How can I prove this??

This is not true. Take any discrete space, then any singleton is open yet not a limit point.

January 20th 2011, 07:57 PM

mathsohard

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove :(

January 20th 2011, 08:05 PM

Drexel28

Quote:

Originally Posted by mathsohard

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove :(

It's clearly untrue in arbitrary spaces, but since this is an advanced calculus book I'm going to take a wild guess and assume you meant and open subset of . If so then this is true if is non-empty. Let and any neighborhood of it, then since is open we know there exists a neighborhood of such that . Note then that is a neighborhood of contained in . Choose some open ball and use the uncountability of open balls in euclidean space to find a point different from in .

January 21st 2011, 03:36 AM

Plato

Quote:

Originally Posted by mathsohard

This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove :(

@mathsohard, you should give us the total context of a question.
I just looked up that question. Not only is it not a general top-space it is a question about subsets of the real number line.